(Occurrences of each digit in a string) Write a method that counts the occurrenc

Question

(Occurrences of each digit in a string) Write a method that counts the occurrences of each digit in a string using the following header:

public static int[] count(String s)

The method counts how many times a digit appears in the string. The return value is an array of ten elements, each of which holds the count for a digit. For example, after executing int[] counts = count("12203AB3"), counts[0] is 1, counts[1] is 1, counts[2] is 2, counts[3] is 2.

Use input and output dialog boxes.

The program should keep running until the user enters “quit”.

And here is my code, but I have trouble to complish "The program should keep running until the user enters “quit”.". I will offer the lifesaver to the one solve this problem.

import javax.swing.JOptionPane;

public class CountDigit{

    public static void main(String[] args){
        String s = JOptionPane.showInputDialog("Enter a 10-digit number, or the word "quit": ");

        int[] count = CountDigits.count(s);   
        for (int i = 0; i < count.length; i++){
            if(count[i] != 0)
                JOptionPane.showMessageDialog(null, "The occurence of " + i + " is " + count[i]);
        }
    }
   
    public static int[] count(String s){
        int[] count = new int[10];
       
        for (int i = 0; i < count.length; i++){
            for(int j = 0; j < s.length(); j++){
                if(s.charAt(j) == 48 +i)
                    count[i]++;
            }
        }
    return count;
    }
   
   
}

Explanation / Answer

Use this as your new main function. public static void main(String[] args) { while(true) { String s = JOptionPane.showInputDialog("Enter a 10-digit number, or the word "quit": "); if(s.equals("quit")) break; int[] count = count(s); for (int i = 0; i

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